In order to pass $10\,\%$ of the main current through a moving coil galvanometer of $99\,\Omega$,the resistance of the required shunt is ............ $\Omega$.

  • A
    $9.9$
  • B
    $10$
  • C
    $11$
  • D
    $9$

Explore More

Similar Questions

The voltmeter has a range of $10 \ V$ and its internal resistance is $50 \ \Omega$. To increase the range of the voltmeter to $15 \ V$,the resistance which is to be connected is:

$A$ galvanometer can be converted into an ammeter by connecting:

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series,showing a full-scale deflection of $30$ divisions. The additional series resistance required to reduce the deflection to $20$ divisions is: (in $Omega$)

Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ For measuring the potential difference across a resistance of $600\,\Omega$,the voltmeter with resistance $1000\,\Omega$ will be preferred over a voltmeter with resistance $4000\,\Omega$.
Reason $R:$ $A$ voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance.
In the light of the above statements,choose the most appropriate answer from the options given below.

Explain the use of a galvanometer as a voltmeter.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo